Question

31. Drosophila females of wild-type appearance but heterozygous for three autosomal traits are mated to males that are homozygous recessive for the three traits.  The three autosomal recessive traits are glassy eyes (g), coal-colored bodies (c), and striped thoraxes (t). The three dominant traits are red eyes (g⁺), gray bodies (c⁺) and solid thoraxes (t⁺) (remember, the wild-type allele has a ⁺ after it).

You examine 1000 progeny of the above cross and they have the following phenotypic classes:    

                        265      g⁺c⁺t⁺

                        275      gct

                                    120          g⁺ct

                                    140          gc⁺t⁺

                                    16             gct⁺

                                    24             g⁺c⁺t

                                    70             gc⁺t

                                    90             g⁺ct⁺

1. Which two phenotypes are the parentals?

2. What is the order of the genes on the chromosome? (in other words, which one is in the middle?)

Map the distances between the three loci.  

Answer 1

Cross: g+ c+ t+/ g c t   X   g c t/ g c t

Questions:

Which two phenotypes are the parentals?

Ans. The phenotype that have the maximum number of progeny are the parental i.e. g+ c+ t+ and g c t.

What is the order of the genes on the chromosome?

The order of the gene can be deduced by looking at the DCO (double crossover) progeny. The DCO progeny are the one having least number of progeny i.e. 16  g c t+ and 24  g+c+t.

The gene whose allele differ in both the phenotype is in the middle i.e. t+ in this case.

So the order of the gene is as follows g_t_c (thoraxes “t” is in the middle)

Map the distances between the three loci.

SCO Region I are (g+c t) and (g c+t+) = 120 + 140 = 260

DCO are (g c t+) and (g+c+t) = 16 + 24 = 40

Total progeny = 1000

Map distance between (g+ and t+) = (SCO Region I + DCO / total progeny) X 100

                                                         = (260+40 / 1000) X 100

                                                         = 30 cM

SCO Region II are (g c+t) and (g+c t+) = 70 + 90 = 160

DCO are (g c t+) and (g+c+t) = 16 + 24 = 40

Total progeny = 1000

Map distance between (t+ and c+) = (SCO Region II + DCO / total progeny) X 100

                                                         = (160+40 / 1000) X 100

                                                         = 20 cM

Distance between g+ and c+ = (distance between g+ and t+) + (distance between t+ and c+)

                                              = 30cM +20cM

                                              = 50cM